Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Give the IUPAC name of each: (a) ; (b) ; (c) ; (d) .
Alkanes , alkenes , alkynes . A homologous series differs by , 14 u. Name = prefixes (branches, in alphabetical order) + parent chain + suffix (principal group). Use the longest chain, with the principal group and the multiple bonds in it. Number to give the lowest locants at the first point of difference: principal group first, then multiple bonds, then branches. Seniority: acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine, alkene, alkyne.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. (a). The longest chain has five carbons. The methyl group is on carbon 2 from the left and on 4 from the right. The name is 2-methylpentane.
Step 2. (b). The longest chain has five carbons, and the two methyl groups sit on the middle carbon. The name is 3,3-dimethylpentane.
Step 3. (c). The double bond is after carbon 2 from the left. The name is pent-2-ene.
Step 4. (d). The alcohol gives the suffix. The OH is on carbon 2 from the right and on carbon 4 from the left. The name is pentan-2-ol.
Answer: 2-methylpentane, 3,3-dimethylpentane, pent-2-ene and pentan-2-ol.
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